Question

**Part A**

The activation energy of a certain reaction is 43.9 kJ/mol . At 28 ∘C , the rate constant is 0.0130s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

**Part B**

Given that the initial rate constant is 0.0130s−1 at an initial temperature of 28 ∘C , what would the rate constant be at a temperature of 100. ∘C for the same reaction described in Part A?

Express your answer with the appropriate units.

Answer #1

we have to use the equation

ln(k2/k1) = (Ea/R) · (1/T1 - 1/T2)

K2/K1 = 2

Ea = activation energy give =43.9 kJ/mol = 43900 J/mol

R =gas constant 8.314 J/mol-K

T1 = 273 + 28 = 301 K

put all these in above equation

ln (2) = 43900/8.314 [1/301 -1/T2]

0.693 = 5280.25 [0.0033 -1/T2]

[0.0033 -1/T2] = 0.693 / 5280.25

[0.0033 -1/T2] = 0.00013

1/T2 = 0.00317

T2 = 1/0.00317

T2 = 315.46 K

partB

use the same equation we need to find out K2

ln(k2/k1) = (Ea/R) · (1/T1 - 1/T2)

ln [K2 / 0.0130] = 43900 / 8.314 [1/301 - 1/373 ]

ln [K2 / 0.0130] = 5280.25 [ 0.0033 - 0.0026]

ln [K2 / 0.0130] = 3.7

[K2 / 0.013] = e^{3.7}

[K2 / 0.013] = 40.4473

K2 = 0.5258 s^{-1}

Part A:
The activation energy of a certain reaction is 50.0 kJ/mol . At
25 ∘C , the rate constant is 0.0110s−1. At
what temperature in degrees Celsius would this reaction go twice as
fast?
Express your answer with the appropriate units.
Part B:
Given that the initial rate constant is 0.0110s−1 at
an initial temperature of 25 ∘C , what would
the rate constant be at a temperature of
140. ∘C for the same reaction described in
Part A?
Express your...

Part A
The activation energy of a certain reaction is 36.4 kJ/mol . At
21 ∘C , the rate constant is 0.0160s−1. At what
temperature in degrees Celsius would this reaction go twice as
fast?
Express your answer with the appropriate units.
Hints
Part B
Given that the initial rate constant is 0.0160s−1 at an initial
temperature of 21 ∘C , what would the rate constant be
at a temperature of 130. ∘C for the same reaction
described in Part A?
Express your...

Part A
The activation energy of a certain reaction is 30.5 kJ/mol . At
30 ∘C , the rate constant is 0.0180s−1. At what
temperature in degrees Celsius would this reaction go twice as
fast?
Express your answer with the appropriate units.
T2 =
Part B
Given that the initial rate constant is 0.0180s−1 at an initial
temperature of 30 ∘C , what would the rate constant be
at a temperature of 170. ∘C for the same reaction
described in Part A?
Express...

Part A:
The activation energy of a certain reaction is 42.8 kJ/mol . At
28 ∘C , the rate constant is 0.0190s−1. At what temperature in
degrees Celsius would this reaction go twice as fast?
Part B:
Given that the initial rate constant is 0.0190s−1 at an initial
temperature of 28 ∘C , what would the rate constant be
at a temperature of 150 ∘C for the same reaction
described in Part A?

A. The activation energy of a certain reaction is 42.0 kJ/mol .
At 29 ∘C , the rate constant is 0.0190s-1. At what
temperature in degrees Celsius would this reaction go twice as
fast? Express your answer with the appropriate units.
B. Given that the initial rate
constant is 0.0190s-1 at an initial temperature of 29 C
, what would the rate constant be at a temperature of
120. ∘C for the same reaction described in Part
A?
Express your answer...

Part A The activation energy of a certain reaction is 30.6
kJ/mol . At 22 ∘C , the rate constant is 0.0110s−1. At what
temperature in degrees Celsius would this reaction go twice as
fast? Express your answer with the appropriate units. Hints T2 = 39
∘C SubmitMy AnswersGive Up Correct Part B Given that the initial
rate constant is 0.0110s−1 at an initial temperature of 22 ∘C ,
what would the rate constant be at a temperature of 170....

Part A: The activation energy of a certain
reaction is 43.5 kJ/mol . At 23 ∘C , the rate constant is
0.0180s−1. At what temperature in degrees Celsius would this
reaction go twice as fast?
Part B: Given that the initial rate constant is
0.0180s−1 at an initial temperature of 23 ∘C , what would the rate
constant be at a temperature of 190. ∘C for the same reaction
described in Part A?

Part A: The activation energy of a certain reaction is 42.3
kJ/mol . At 29 ∘C , the rate constant is 0.0170s−1 . At what
temperature in degrees Celsius would this reaction go twice as
fast?
Part B: Given that the initial rate constant is 0.0170s−1 at an
initial temperature of 29 ∘C , what would the rate constant be at a
temperature of 120. ∘C for the same reaction described in Part
A?

Part A
The activation energy of a certain reaction is 48.4 kJ/mol . At
24 ∘C , the rate constant is 0.0190s−1. At what
temperature in degrees Celsius would this reaction go twice as
fast
T2 =
35 ∘C
Part B
Given that the initial rate constant is 0.0190s−1 at an initial
temperature of 24 ∘C , what would the rate constant be
at a temperature of 190. ∘C for the same reaction
described in Part A?
k2 =?

The activation energy of a certain reaction is 35.1 kJ/mol . At
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Given that the initial rate constant is 0.0160s−1 at an initial
temperature of 25 ∘C , what would the rate constant be
at a temperature of 200. ∘C for the same reaction
described in Part A?

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